Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{k(5k + 6)}{10} \times \dfrac{-4}{10k^2 + 12k} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ k(5k + 6) \times -4 } { 10 \times (10k^2 + 12k) } $ $ z = \dfrac {-4 \times k(5k + 6)} {10 \times 2k(5k + 6)} $ $ z = \dfrac{-4k(5k + 6)}{20k(5k + 6)} $ We can cancel the $5k + 6$ so long as $5k + 6 \neq 0$ Therefore $k \neq -\dfrac{6}{5}$ $z = \dfrac{-4k \cancel{(5k + 6})}{20k \cancel{(5k + 6)}} = -\dfrac{4k}{20k} = -\dfrac{1}{5} $